groups-rings-fields

[twilit pawn]

weird spacing...

anyway, we can assume choices have been made so $span B_1$ and $span B_2$ don't overlap

[cloud walrus]

doubledual

[twilit pawn]

and we have remaining vectors $c_l, \dots, c_n$ which are in the span of neither $B_1$ nor $B_2$.

[cloud walrus]

doubledual

[twilit pawn]

so we take $c_l$ and add it to $B_1$.

[cloud walrus]

doubledual

[twilit pawn]

If this puts any new $c_i$ in the span of $B_1$, we add those to $B_2$.

[cloud walrus]

doubledual

[twilit pawn]

now that might force us to add stuff to B_1, so we alternate back and forth as much as needed

until we return to the start of the algorithm

I believe that alternating as I described prevents us from having issues

i need to think a little harder there

[final pasture]

(note: this isn't even clear that this is always possible without losing linear independance, that definitely needs to be proved)

[twilit pawn]

agreed

[final pasture]

(this is a surprisingly hard exercises given this is an exercise from the first chapter of a linalg textbook lol, the other weren't as hard )

[twilit pawn]

ok here is why I think alternating works

so we add $c_l$ to $B_1$

[cloud walrus]

doubledual

[twilit pawn]

so we then have some c's which are in the span of B1, but not B2

everything else is in the span of neither

we add those cs to B2

so now of the leftover vectors, some are in the span of B2 and not B1, and the rest are in neither span

so as you alternate, you always have this property, no c vector is ever in both spans

at each stage as you add vectors, you preserve linear independence for this reason

so it works

hmmm i think i see a snag actually

i can't really add groups of vectors to the lists

the span of the stuff I add might intersect the span of the Bi we add it to

even if no vector is already in the span

i think we need to do something close to this, but one vector at a time

[final pasture]

(I'm still here btw, I'm trying stuff but I get nothing that seems worth of interest )

[twilit pawn]

(same lol)

here's an interesting thought

[final pasture]

👀

[twilit pawn]

this reduces to the following problem

if you have two bases, can you always swap a given vector in the first basis with some vector in the second basis, and still have both be bases?

[final pasture]

how does this reduce to this problem ? 🤔

[twilit pawn]

if you can, you just do that m times

[final pasture]

no ?

[twilit pawn]

between the B and C basis

[final pasture]

imagine you swap the same vectors m times

[twilit pawn]

no so my statement is this

you have a basis b1, ..., b_n

and

c1, ..., cn

is there a ci

so ci, b2, ..., bn is a basis

and b1, c1, ..., cn is a basis

(no ci in the second list)

[final pasture]

yeah I get that

but imagine (a1, a2) and (b1, b2) are our basis

[twilit pawn]

oh wait i get your objection

yeah sorry 😦

[final pasture]

But actually

I don't even see how to solve the m = 1 case

[twilit pawn]

something like this may still be helpful though

[final pasture]

so maybe this is worth thinking about it

or maybe

wait

just a random thought: is the m = n-1 case easier to solve ?

nvm, probably not

[twilit pawn]

basically the same case lol

[final pasture]

I was thinking about some induction but nvm

[twilit pawn]

so here's one thing you can do

if you have b2, ..., bn

at least one vector in c1, ..., cn is not in the span of those

so call that vector ci

so ci, b2, ..., bn is a basis

is b1, c vectors a basis?

[final pasture]

my first counter-example works here, doesn't it ?

[twilit pawn]

doesn't seem like it has to be a basis...

yeah i agree

so yeah how would we even switch one vector lol

[final pasture]

btw, not sure if you have seen this

[twilit pawn]

i did, i didn't have an ideas based off of it but its a reasonable approach

[final pasture]

Phrased like this, the problem looks a bit easier, even though I'm unable to prove it

oh also, how about induction on the size of the family ? I didn't even consider it yet

[twilit pawn]

hmm that might be a better approach

you can probably do some sort of subspace analysis that way

small amount of cases that might make it viable

can we come up with an argument in the 2d case?

[final pasture]

Let's try ?

I think it's bruteforceable

a1 a2 and b1 b2 basis.
If a1 not colinear to b1 and a2 not colinear to b2, we're done.
Else, if a1 colinear to b1 then a2 can't be colinear to b1 and similarly b2 can't be colinear to a1

And same thing in the remaining case

[twilit pawn]

so an argument like this with induction, but replace colinear with more fancy words???

[final pasture]

idk, how would you do that ?

maybe let's try the 3d case

[twilit pawn]

seems good

[final pasture]

brb

am back

[twilit pawn]

im in a zoom seminar now, but ill let you know if i come up with anything

[final pasture]

If anyone wants to take a look at the problem:
https://media.discordapp.net/attachments/496784958430380033/845359901396172800/406586081627078666.png?width=1440&height=560

We only solved the n = 2 case

[final pasture]

(btw thanks for the help ! )

[next obsidian]

Should that last c_n also be c_sigma(n)?

Also, I have no idea how to do this

[final pasture]

it should be c_sigma(n) yeah

[final pasture]

Neither do I

[nova plank]

What french horror novel is this from?

[final pasture]

it's from roman's Advanced Linear Algebra

[nova plank]

Ah

[final pasture]

one of the last exercises of the first chapter

Definitely a legal link
http://matematicas.uis.edu.co/sites/default/files/paginas/archivos/Advanced Linear Algebra - Steven Roman.pdf

[next obsidian]

I can only say

Grab some bases of R^3

And see why it’s true lol

Look at a few and maybe some pattern emerges

¯_(ツ)_/¯

[final pasture]

That's currently my plan, yeah

[next obsidian]

Well okay so like

If you take b1 through bm

And quotient out by them

Then c1 through cn span the quotient

I guess their like, quotients do

So you can grab n - m of them

And they’ll span V /<b1,...,bm>

But that’s the same thing as <bm+1,...,bn>

So this at least gives you some sigma such that
(b1,...,bm,c_sigma(m+1),...,c_sigma(n)) is a basis

[final pasture]

oooh

[next obsidian]

You’d need to show the other one works too

[final pasture]

ok actually wait

[twilit pawn]

this is a very elegant way to get one basis

idk if the other works lol

[final pasture]

lemme check that my first counter-example doesn't apply here too

[next obsidian]

Actually it totally does

Because the c_i that died in that quotient

Lie inside of <b1,...,bm>

But that’s m things linearly independent in there

So <b1,...,bm> = <leftover m c_i>

[twilit pawn]

do we know we have m things in there?

[next obsidian]

Yeah

Because you the images of c1 through cn inside of

V/<b1,...,bm>

[twilit pawn]

so n-m of them form a basis

I agree

[next obsidian]

That quotient is n-m dimensional

[twilit pawn]

but I don't think m have to die

[next obsidian]

So there’s m that have to die

[twilit pawn]

lol

[next obsidian]

Uhhh

[twilit pawn]

they could be linearly dependent right?

[next obsidian]

I mean

So either they live in <b1,...,bm>

Or in the complement

If they were in the complement

[twilit pawn]

R^2 / (1,0)

look at the basis (1,1), (1,-1)

[next obsidian]

you’d have n - m + 1 thinfs in that n - m dimensional vector space

[twilit pawn]

neither die

[next obsidian]

🤔

[twilit pawn]

either vector will project to a non-zero vector

they aren't LI after projection

they are definitely spanning, so you can get a basis that way

[next obsidian]

Okay well here’s a different way to see it I guess

So maybe that’s a bit off but

I think this works like

Hngg

Tfw no paper on hand

[twilit pawn]

before you keep going, let me try to apply shika's counterexample here

B = (1,0), (0,1)

C = (1,0), (1,1)

say you quotient out by (0,1)

both vectors in C are mapping to 1

so you could take either as your swap candidate

but swapping (1,0) would not work

[final pasture]

Isn't chmonkey idea actually just your first idea phrased in a fancier way ?

[twilit pawn]

yes

lol

[next obsidian]

Rip

You have to control it harder

[twilit pawn]

its a much better phrasing

of my idea which doesn't work

[next obsidian]

Gggg

Honestly

[twilit pawn]

the quotienting idea is pretty interesting though

[next obsidian]

I kinda feel like maybe if you write both bases in terms of each other

Then you can do some bullshit

[final pasture]

I tried to do so

[next obsidian]

So obviously you can lift an arbitrary basis of the quotient up

[final pasture]

https://cdn.discordapp.com/attachments/496784958430380033/845362728582774784/406586081627078666.png

[next obsidian]

Since your computation shows that it won’t work

What did you do in the dim 2 case?

[final pasture]

just pure bash

[next obsidian]

Bashing out what?

The matrix?

[final pasture]

no, there's actually only 3 cases to handle in the dim 2 case

lemme write out the argument here

[next obsidian]

Or you can link to it

I can just read that

[final pasture]

so let (a1, a2) and (b1, b2) be basis.
If a1 isn't colinear to b2 and b1 isn't colinear to a2, then we can swap a1 and b1.

Else, if a1 is colinear to b2, then a2 can't be colinear to b2, else a1 and a2 would be colinear.
Similarly, b1 can't be colinear to a1, else b1 and b2 would be colinear.
So swapping a1 and b2 work.

Same thing in the last case (a2 colinear to b1)

[next obsidian]

Oof

[final pasture]

I could've linked it but I wanted to rewrite it just to be sure I didn't messed up somewhere

[next obsidian]

Okay wait am I dumb or...

Okay this is a more abstract way to think about this but

You can define maps out of a vector space

By just saying where the basis elements go

[final pasture]

yeah

[next obsidian]

So like

Hmmm

Okay nvm I see the issue with this

Lol

[final pasture]

what was your idea ?

[next obsidian]

Just map shit and swap one at a time, but there’s no way to know that still forms a basis

But maybe you can like

Say if no such choice works... then....

Okay so

If you had (b1,...,bn) a basis

And (c1,...,cn) a basis

What if you had like
(ci,b2,...,bn) not a basis for all n

I feel like you’d want to say that then

[final pasture]

this isn't possible

[next obsidian]

All ci are in the span of b1 through bn

Yeah

But this implies something like

So like

You could try to say “none of them span V, so something bad”

Or something like “there’s a linear dependence, so something bad”

[final pasture]

I think what you want to assume is:
(ci,b2,...,bn) and (b1, c2, .., cn) are not simultaneously basis for all i ?

[next obsidian]

Like my idea is that somehow this implies either

[final pasture]

no ?

[next obsidian]

(c1,...,cn) doesnt span

Or isnt linearly independent

[final pasture]

yes

[next obsidian]

Because I think if (ci,b2,...,bn) is a basis

(c1,...,b1,...,cn) should be

For... reason

[final pasture]

nope

same example than before

[next obsidian]

But like

If (ci,b2,...,bn) is a basis

[final pasture]

A = (1,0), (0,1)
B = (1,0), (1,1)
swap out the (0,1) vector of A with the (1,0) of B

[next obsidian]

Tfw

I felt like the fact a certain map is an isomorphism says it should be so...

Oh I guess the inverse didnt look how I wanted

Ffffff

I should get out of the 🛁 and grab some paper

[final pasture]

[nova plank]

Damn, most of my math help happens while I'm in the bath too.

[final pasture]

go in the bath and help me, then

[nova plank]

😂 😂

[next obsidian]

Okay so

I think I have the literal shittiest proof ever

But I don’t want to say it because it’s so overkill

[twilit pawn]

i have a possibly less shitty proof

possibly

[final pasture]

Now I want to see it

say it please

[nova plank]

Chmonkey proved a millenium prize problem to solve this problem.

[twilit pawn]

if the overkill one is short just do it lol

"by fermat's last theorem...."

[final pasture]

and go ahead share your potential proof too, doubledual

[final pasture]

"Ok so assuming RH.."

[twilit pawn]

so let B1 = b_1, ..., bm

B2 = bm+1, ..., bn

we're going to come up with an algorithm to sort the cs into these two groups

so first off, each c is in span B1 but not B2, or B2 not B1, or neither

we start with c1

[final pasture]

yeah, because B1 and B2 are supplementary

[twilit pawn]

if it's in the span of B1, we're forced to put it in B2, similarly if it's in span B2

now I claim that if we add it to B1 (say), each c still vector still satisfies the trichotomy

if some ci was in the span of B1, and B2, that ci must have been in the span of B2 originally

sorry i think i lost the idea D:

there was a dimension idea in here

[next obsidian]

Oof, my idea doesn’t pan out either I think

[twilit pawn]

like i was going to get too many LI vectors somehow

[nova plank]

let's ask on chegg

[next obsidian]

I wanted to use the fact that if P,P’ are projective modules surjecting onto the same thing via say f,f’ then kerf + P’ = ker f’ + P

Lmfao

[twilit pawn]

ok let me try and assume that we can sort the ones which are forced to be sortd

because otherwise the problem is false

just to see if I can deal with the case of c vectors in neither span

yeah ok so if a c vector is in neither span

you add it to B1

the only thing to ask is if a vector in span B2 can now be in span B1

oh maybe i should keep track of the c vectors and play with direct summing of vector spaces...

[next obsidian]

I may have a legit proof now holy crap

I just need to see if it looks like crap or not lmao

Okay so...

Suppose that no c_i are in <b_1,...,b_m> first

then {b_1,...,b_m} along with any n - m c_i forms a basis of V

because all the c_i are non-zero inside the quotient

does that seem good?

[twilit pawn]

no

[final pasture]

no

[twilit pawn]

lol

[next obsidian]

wut

[final pasture]

xD

[twilit pawn]

yeah they don't all have to be non-zero

[next obsidian]

I know

But I said suppose

[twilit pawn]

ohhhh wait sorry

[next obsidian]

because I have a proof for when it is

[twilit pawn]

you did do that

[next obsidian]

I think

[twilit pawn]

ok i agree then

maybe?

[next obsidian]

I mean if they don't live in there

then all c_i are still nonzero in V/<b_1,...,b_m>

[final pasture]

what tells you that you can't have c_i written as c_j + b with b € Span {b1, .., bm} ?

[next obsidian]

so any n - m of them forms a basis of it

am I right...

maybe I'm still wrong

F

well whatever, I think you can prove the case where some c_i is in <b_1,...,b_m> by induction

which at least is a step forward

[twilit pawn]

ok i have a partial proof

where the missing step must work if the problem is true

[nova plank]

Why is this problem so wild?

[twilit pawn]

idk lmao

[nova plank]

I looked in the book now

And it gives a hint

[twilit pawn]

hmmmmmm

[next obsidian]

tfw

[nova plank]

New edition added a hint because it was too hard, maybe 😂

[twilit pawn]

yeah lmao

ok just to get it out there

I have my B1 and B2 from before

every c vector is in the span of at most 1

[final pasture]

can you share the pdf of the new edition please ?

[nova plank]

No, that is piracy

[twilit pawn]

first suppose we can add c vectors which are in the span of one list to the other list

[final pasture]

meeh

[twilit pawn]

and maintain the property that everything in the span of at most 1 list

(I believe this must be possible for the problem to be true)

do this until every remaining vector is in the span of neither list

if you add such a vector to one list, every remaining vector is now in the span of at most 1 list

so we've maintained that property

so it's going to be possible, if we can show the property is maintained when the vector is in the span of one list

that make sense?

so really we reduce it to the following case:

I have c1 in the span of B2 and not B1

I add c1 to B1

c2 was in the span of B2 already

can c2 be in the span of B1?

oh it definitely can't if B1 and B2 are the original basis, because that would break linear independence

but it's less clear if we've added other vectors to them

[final pasture]

funnily enough the only exercise that got an hint added is this one

[twilit pawn]

hahahaha

[nova plank]

😂

[final pasture]

(among the exercises of the first chapter, I mean)

[nova plank]

Idk how to use the hint anyway

[twilit pawn]

ok let me try this approach I cooked up

so what I think this reduces to:

[nova plank]

I have forgotten too much LA

[final pasture]

I don't know how to prove the hint anyway, so I don't want to use it

[twilit pawn]

B1 = <b1, ..., bm>

[final pasture]

[twilit pawn]

B2 = <bm+1, ..., bn>

[nova plank]

Lol, the hint might be there for a reason. I would use it if you can figure out how.

[twilit pawn]

and let's say we've made choices for the first k c vectors

so we have B1 and C1 direct summing, and B2 and C2 direct summing

and moreover, every remaining vector is in the span of at most one of these sums

so we consider the next vector c on our list, and suppose c is in B2 + C2

so we add c to C1

we need to verify that if c' is in the span of B2+C2, that c' is not now also in the span of B1+C1

[final pasture]

idk how to use it anyway

[twilit pawn]

so let's say for contradiction $c' = b_2+c_2 = b_1 + c_1 + \lambda c$

[cloud walrus]

doubledual

[twilit pawn]

bi in Bi and ci in Ci, lambda a scalar

[nova plank]

I think it can after you add a vector to it, btw.

[twilit pawn]

oh wait

if c' is in B1 + C1

then c would have to be in B1 + C1

which we assume is not the case

ok so let me just make this crystal clear

we have sorted some of the vectors into these sets C1 and C2

the next vector to sort is c

c is in B2+C2, but not B1+C1

since we have the inductive hypothesis that every vector in the c basis is in at most one of these direct sums

so we add c to C1

we need to check the inductive hypothesis is still holding

if c' is an unsorted basis vector

if it was originally in neither span we're fine, at most it's in one span

if it was in B1+C1 we're also fine

so we have to consider the case the c' is in B2 + C2 but not B1 + C1

if c' were in B1 + C1 + <c>

[final pasture]

(I totally lost track of what you're doing doubledual )

[twilit pawn]

sorry lmao

if i convince myself of this i will backtrack

[final pasture]

thanks

[twilit pawn]

so if c' = b1+c1+ lambda c = b2+c2

[latent anvil]

Is this the problem yall are discussing?

[twilit pawn]

ok i think i confused myself above and this isn't working

[twilit pawn]

yeah

[final pasture]

yes

[twilit pawn]

ok i give up this problem has beat me

[latent anvil]

hmm

So my instinct is

Take the bi to be the standard basis for k^n

The other basis is some matrix A in GL_n(k)

We want to show you can pick m of the rows such that the matrix with the standard basis vectors and then those rows is invertible, and the same for their complements

What's the determinant of a matrix which is like e1,...,em,c1,...,cr?

[final pasture]

https://cdn.discordapp.com/attachments/496784958430380033/845362728582774784/406586081627078666.png

[latent anvil]

Can you do like, cofactor expansion ?

[final pasture]

that's what I was saying here, right ?

[latent anvil]

Yup, this is what I'm thinking ajout

So what's the determinant of a generic matrix whose first m rows are the same as the identity

[final pasture]

huh wait

I think I know something that may apply here

[latent anvil]

You should be able to do cofactor expansion

But I haven't done that in years so it's taking me a minute haha

I think you get a sign times a minor of the other matrix

[final pasture]

wouldn't such a matrix be block triangular ?

[latent anvil]

No

[final pasture]

oh wait yeah no nvm

[latent anvil]

An example is
1 0 0 0
0 1 0 0
2 3 4 4
7 1 0 3

But the determinant of this is ± the determinant of
4 4
0 3

I think it'll always be + but uh

It's been a minute

[final pasture]

lol

[latent anvil]

Do you see what I'm saying tho?

[final pasture]

well isn't this block triangular ?

[latent anvil]

This?

[final pasture]

yes

[latent anvil]

Wait maybe I misunderstood what block triangular is

You can split it up into blocks where the upper right one is zero

the blocks will not have the same dimensions

[final pasture]

we don't care about that

[latent anvil]

So say your starting basis is c1,...,cr and let A be the matrix with them as rows. Then e1,...,em, cσ(1),...,cσ(n-m) is a basis iff the bottom right minor of A corresponding to cσ(1),...,cσ(n-m) is nonzero

This is what my computation was doing

[final pasture]

you still have that the determinant of the whole matrix is the product of determinants of the diagonal matrices

[latent anvil]

Ah sure

[final pasture]

even if the blocks are rectangular

[latent anvil]

I was thinking about this via cofactor expansion

But the block triangular thing should apply too

So our result can be rephrased as something like

Say you have an invertible matrix A

Find two minors of appropriate size/position which are nonzero

[final pasture]

also

did you see the hint ?

https://media.discordapp.net/attachments/496784958430380033/845401672724643880/Screenshot_20210521-215249_Adobe_Acrobat.jpg

the new edition of the textbook contains an hint

[latent anvil]

No I did not

but uh

I think this is what I was saying?

Right?

[final pasture]

idk tbh lol, I'm kinda confused

[latent anvil]

Sure so

Think about it like this

Say you found a permutation which works

The collection e1,...,em, cσ(1),...,cσ(n-m) being a basis means something in terms of matrices

Yeah?

It means the matrix with those rows has invertible determinant

[final pasture]

yeah

[latent anvil]

but we calculated the determinant of that matrix

You has the block triangular thing

So it's actually the determinant of this weird submatrix of the matrix with rows c1,...,cn

[final pasture]

we're expressing the rows in the (e_i) basis, right ?

[latent anvil]

Or not quite sorry, if you take that matrix with rows c1,...,cn and then permute the rows by σ it's the determinant of a submatrix

Yup!

Ah sorry I hint and I got flipped around

The problem says b1,...,bm,cσ(m+1),...,cσ(n)

When I said bm,cσ(1),...

It doesn't really matter but this is why they have the bottom right (n-m)×(n-m) submatrix and I would have the top right (n-m)×(n-m) submatrix

So let's stick with their convention

[final pasture]

ok yeah

[latent anvil]

So now it's a problem of matrices

Oh um wait

The hint says we can use this

So we've solved to the hint's standards

I'm still not sure why it's true though

[final pasture]

huh

I'm kinda confused about what we solved lol

ig the exercise

but like

for the moment didn't you just reformulate the problem in term of matrices ?

I think I got this part, since it's pretty much what I was trying to do

[latent anvil]

Yeah

[latent anvil]

The hint says "you may assume..."

[final pasture]

but then, what do you do after that, and how do you use the hint ?

[latent anvil]

We reduced it to exactly what the hint says

[final pasture]

how so ?

[latent anvil]

We showed it suffices to show that there's a permutation of the matrix with the rows c1,...,cn where the appropriate minors are nonzero

Which is what the hint says we can assume

Okay so let me go from the top

wtf

[final pasture]

I spent way too much time thinking about this exercise, I think I'm lacking brainpower, but I don't think I understand at all where we did that

[latent anvil]

This server doesn't show up on my laptop

Okay it's here now

I'll explain

[final pasture]

ok thanks

[latent anvil]

say you have bases b1,...,bn and c1,...,cn

write $c_i = \sum_{j=1}^n a_j^i b_j$ and let $A$ be the matrix with rows $a_1^i,\ldots,a_n^i$

[cloud walrus]

shamrock

[latent anvil]

sorry einstein summation notation is a force of habit

the hint tells us we can find a permutation $\sigma \in S_n$ where if $B$ is the matrix where we permute the rows of $A$ according to $\sigma$ then the upper left $m\times m$-submatrix and lower right $(n-m)\times(n-m)$-submatrix of $B$ are invertible

[cloud walrus]

shamrock

[latent anvil]

does that sound right to you?

I'm just applying the hint to A here

[final pasture]

so basically C's matrix in the basis B ?

[latent anvil]

Yup

[final pasture]

I'm totally braindead rn, that'll take time sorry

[latent anvil]

I think it might be transposed possibly

but that's the idea

A is invertible since its rows are a basis

[final pasture]

ok yeah so I agree with that

[latent anvil]

cool so I claim that $b_1,\ldots,b_m,c_{\sigma(m+1)},\ldots,c_{\sigma(n)}$ and $c_{\sigma(1)},\ldots,c_{\sigma(m)},b_{m+1},\ldots,b_n$ are bases for the space

[cloud walrus]

shamrock

[latent anvil]

it suffices to show that the matrices with these vectors (expressed in the basis $b_1,\ldots,b_n$) as rows are invertible, yeah?

[cloud walrus]

shamrock

[final pasture]

yeah

oh and wait

[latent anvil]

equivalently, the determinant is invertible

[final pasture]

yeah

ok I think I see where this is going

[latent anvil]

and we already calculated that determinant!

[final pasture]

and I think I'd be able to finish from here

yeah right

[latent anvil]

👍

[final pasture]

thanks a lot

[nova plank]

Damn, even I get it now go shamrock

[latent anvil]

I don't see how to prove the hint though, sorry

[final pasture]

I'm a bit too braindead rn, but I'll try to write up the full solution with all the details tomorrow

[nova plank]

And try to prove the hint 😅

[final pasture]

as for the hint, I think a friend of mine got a proof, if I really am unable to prove it, I'll ask him

[latent anvil]

oh sick

i was about to ask twitter

also i wonder if there's geometry to this

but I think several people failed to see it before me

so it might just be ugly

[final pasture]

ok so actually I think I don't want to prove the hint

(that's what the general laplace expansion theorem states, I think)

http://www.imvibl.org/imvibl/buletin/bultetin_15_2008/buletin_15_2008_5_7.pdf

[latent anvil]

huh

I was thinking of like, the sum of products times a sign

what's the conjugation?

[final pasture]

I think I'll just assume the hint is true, that looks ugly

[final pasture]

the \bar{D} ?

[latent anvil]

yeah

you know what

i just don't awnt to think about this

this notation is annoying

oh wait

are you over a ring or a field?

like, does det A ≠ 0 imply A is invertible?

if so I think this might actually be obvious from the general formula

@final pasture

[final pasture]

I'm over a field @latent anvil

[latent anvil]

ahhh

then it is obvious

[final pasture]

[latent anvil]

this sum is nonzero

right?

so some term in the sum must be nonzero

(as 0 + 0 + ... + 0 = 0)

[final pasture]

yeah right

[latent anvil]

sorry so take i1 = 1, i2 = 2, ..., ir = r

then $\text{sign} \cdot D\left[\overset{i_1\ i_2 \ldots i_r}{j_1\ j_2 \ldots j_r}\right] \overline{D}\left[\overset{i_1\ i_2 \ldots i_r}{j_1\ j_2 \ldots j_r}\right] \neq 0$

[cloud walrus]

shamrock

[latent anvil]

let $\sigma(\ell) = j_\ell$

[cloud walrus]

shamrock

[latent anvil]

then $D\left[\overset{i_1\ i_2 \ldots i_r}{j_1\ j_2 \ldots j_r}\right]$ is the determinant of the upper left $r \times r$ of the matrix you get after permuting the columns of your matrix by $\sigma$

[cloud walrus]

shamrock

[latent anvil]

and the conjugate thingy is the determinant of the bottom n-r by n-r submatrix

so if their product is nonzero, this means they're both invertible, as desired

This is definitely a weird problem

[final pasture]

it's like

impossible to do it

without the hint

and it's not like we're supposed to find the hint

the problem is way before the chapter on the determinant

I'm kinda salty tbh, I spent weeks on it, and it's like, kinda obvious with the hint (I'm confident it would've taken me at most a few hours with the hint )

[latent anvil]

Yeah definitely

Like both of us were able to reduce to it

But then to prove the remaining claim you need this weird expansion formula for the determinant

[final pasture]

maybe this weird formula isn't necessary

i'll google that quickly

[latent anvil]

Well it is pretty easy to prove the hint with this formula, the notation is just ass

Is this for a class or just a book you're working through?

[final pasture]

just for a book I'm working through, I don't have linalg classes yet, I'm still in hs

[final pasture]

(also thanks @next obsidian, I think I forgot to thank you for trying to help )

[next obsidian]

oh, no problem haha

:)

[vestal snow]

Let F be a presheaf and let F^sh be its sheafification. By construction, F^sh(U) is the set of compatible germs on F(U). We know that the natural map F(U) --> \prod_{p\in U}F_p is injective with image equal to the set of compatible germs on F(U). Doesn't this imply that F and F^sh are isomorphic?

[latent anvil]

I don't think that's the image

I think you're thinking that the sheaf image of F -> product of stalks sheaf is F^sh

But the sections of the image sheaf over U aren't the image of the map on sections

Wait, it's also not necessarily injective

[vestal snow]

Oh wait you're right

[latent anvil]

Take the presheaf on a discrete two point space which is Z globally and 0 on all other pens

[vestal snow]

It's injective when F was a sheaf

In which case F is ismorphic to F^sh

[latent anvil]

Not quite

It's injective when F is a separated presheaf

So being sheaf has two properties yeah?

Gluing exists and gluing (if it exists) is unique

The second one makes you a separated presheaf

[vestal snow]

Got it

[latent anvil]

An example of a separated presheaf whcih isn't a sheaf is any presheaf of functions, eg the presheaf of bounded functions

[vestal snow]

Prove that the construction for F^sh satisfies the universal property. Let f:F --> G be a map of presheaves where G is a sheaf. Then this induces a map f':F^sh --> G^sh. However, the natural map n_G: G --> G^sh is an isomorphism as G is a sheaf. Thus, n^-1 \circ f^sh is the required map. How do I prove uniqueness?

One way is to use that a presheaf morphism from a presheaf to a sheaf on U is determined by its stalks on points of U

Is there some other way to do it without using stalks

[next obsidian]

wut

separated presheaf isn't a presheaf

[next obsidian]

yeah, you construct the sheafification in two steps without using stalks

you have to do this when working with sheaves on a site

In the case of the sheafifcation you're probably familiar with where the sheaficiation is just... built up by the stalks, I don't think you can like...

F^sh is built up by a bunch of stalks, idk how you'd do it without stalks

[latent anvil]

Should've been "isn't a sheaf"

[next obsidian]

:O

[latent anvil]

Fuck

Wrong reply lmao

[next obsidian]

I mean

[latent anvil]

Should've been "isn't a sheaf"

[next obsidian]

I knew what you were replying to haha

TFW

[latent anvil]

Ah so

I transposed sheaf and presheaf

It was supposed to be "presheaf of functions"

this is still not correct as written, I mean any non sheaf presheaf of functions

But whatever lol

[next obsidian]

kekw

I have decided

I have worked with sheaves enough to be able to use them

but i am not nearly arrogant enough to claim I understand them

Sheaves are constantly my struggle with AG lol, they're so fucking weird

[vestal snow]

Are schemes nicer than sheaves?

[next obsidian]

I mean they have a sheaf attached to them, so strictly, no I suppose

but I don't often get tripped up by the scheme bit, it's doing all these weird things with verifying equalities of things on sheaves

and sheaves of modules being odd, blah blah blah

you want to know all these facts which are intuitive, like yes of course tensoring does blank but then you try to verify it and it's just like wtf

[vestal snow]

Wow the notation for sheafification related stuff sucks

Explicitly writing down stalks of the sheafification is harder than the problem itself

[gritty adder]

yeah you ideally don't want to refer to the sheafification at all by just checking stalks of the original map

[vestal snow]

That's what I'm proving

[gritty adder]

rip

[vestal snow]

That the stalks are isomorphic

I would share a picture of my proof but it is very ugly

[sturdy marsh]

sheafifification

[next obsidian]

I think that it's not hard to show that the stalks are isomorphic. You want to look at what F^sh looks like on a small open nbd U of x

F^sh(U) on a small nbd of x should look like a tuple which basically is just (s_p) over all p in U

and when you go more local it should stay looking like that, so I'm pretty sure the maps F(U) -> F^sh(U) given by sending s in F(U) to (s_p) in F^sh(U) are an isomorphism

and since the colimit can be computed locally in this sense, the induced map on the colimits should be an iso

I think this is right?

I guess this shows you that the maps F(U) -> F^sh(U) become surjective, but injectivity is a bit trickier maybe? I think it's not true in general that if s_p = t_p for all p in U, that s = t in a presheaf

Or here's a better way to see this maybe, consider an element s_x in F^sh_x. This can be represented by something like <s,U>, then on some nbd V of x we know that s(q) = t_q for all q in U, with t in F(V)

you can define a map F^sh_x -> F_x sending s_x to t_x, then you can show this is an inverse to the obvious map F_x -> F^sh_x, you have to show that this is well-defined though

but this I think just kinda follows by taking intersections of the various open nbds

Overall I guess "not hard to show that the stalks are isomorphic". Seeing that they should be the same isn't too hard I think by that condition that any function needs to be "locally constant" but proving it is a bit harder

[uncut girder]

@next obsidian sheaves are just covering spaces viewed from below

[bronze talon]

hey everyone just a quick question

for a field k and a finite group G, what does k^|G| denote?

[delicate bloom]

I'd assume since |G| is just a number, it's the set of all elements in k raised to the |G| power

[bronze talon]

oh right holy shit lol

[gritty adder]

I would guess that it's the vector space over k of dimension |G|

[bronze talon]

neck deep in category theory and rep theory rn.. was parsing it as some exotic algebra

[delicate bloom]

yeah I like Kedar's interpretation better

what I'm talking about would be more likely to me if it was (k^x)^|G| I think

[gritty adder]

In the context of rep theory you're probably referring to the vector space of functions G -> k (which is the same thing)

[bronze talon]

by the way has anyone had success going from an MA in pure math to a PhD in philosophy

i don't wanna do this anymore 😄

[vestal snow]

I'm trying to read this paper and I wanted to know what parts of AG I would need to understand it and apply similar techniques to a related research problem

Perhaps a better thing to ask would be what all can I skip?

If someone could look at this list of chapters (not necessarily individual sections) and tell me which I should skip for now, that would be great

For example, I think I should be able to skip the chapter on intersection theory

[dusk summit]

So I'm trying to read aluffi rn and his first chapter on ring theory introduces a little bit of homological algebra. However, to be honest, I have zero idea of what homological algebra is- any explanations of what it is on the internet are far beyond my level. Does anyone know of any undergraduate-level explanations of what homological algebra "is", and why we want to do it?

[chilly ocean]

I had sort of the same question as you, fwiw supposedly rotman has an introductory book on homological algebra, although I haven't read it

[wind steeple]

imo if you don't understand anything you should read for a book presenting more elementary algebraic geometry. Altought I find this one very good and I don't see any chapter you could skip (expect the one marked with a star). You should read all the chapters

[compact needle]

It's a bit hard to say. Vakil's book definitely is not a very quick path to what you want to know. There is a lot more information than you strictly need, but you'll want to know things from every chapter other than possibly 17 and 20 (if you need them, you can black box the results in 17.4). And even then, I don't think Vakil talks at all about the Cartier operator.

[compact needle]

Honestly, I found homological algebra to be pretty mystifying until I learned some algebraic topology.

[maiden ocean]

Do you know what a functor is?

I think the short answer from your context is that exact sequences are very useful computationally because if you know about some terms in the sequence it lets you get information about the others: e.g if we have A -> B -> C -> D exact then A = D = 0 implies B iso C and things like that

We call a functor F exact if it sends exact sequences to exact sequences, e.g ... -> A -> B -> C -> ... exact implies ... -> F(A) -> F(B) -> F(C) -> ... exact

[dusk summit]

Thanks for the answers everyone

@maiden ocean when do exact sequences come up? Aluffi says that a lot of the time modules will appear in "sequences" but idk when that occurs

[maiden ocean]

I think the most common example comes from algebraic topology

Actually well the most common example would be quotients

if H is a subgroup of G then 0 -> H -> G -> G/H -> 0 is exact for example

[rustic crown]

(normal)

[maiden ocean]

But beyond that, like, homology is a good example

[rustic crown]

(but since you used 0, ig you're working in Ab?)

[maiden ocean]

No i was just being sloppy lol

If I have a topological space X i have a bunch of functors H_n, homology functors, which I guess you can think of as capturing "n dimensional data" in some sense

dw about the details

anyway for nice subspaces A of X then we can define relative homology H_n(X, A) which more or less "ignores A", the details of the construction dont matter but just think of it as kinda like a quotient

anyway we get a long exact sequence ... -> H_n(A) -> H_n(X) -> H_n(X, A) -> H_n-1(A) -> H_n-1(X) -> H_n-1(X, A) -> ... @dusk summit like this

or heres another: say I and J are two ideals of a ring A

[cloud walrus]

Moth In Shambles

[maiden ocean]

its kinda hard to even pick just a few examples cause theyre like

extremely common

[dusk summit]

I think I can kind of see now, thanks for the examples

[maiden ocean]

!!! wait

thats supposed to be the direct sum of I and J not the tensor product

[dusk summit]

yeah i was wondering what that was

So what new information do the exact sequences give?

For example in the one above with normal subgroups

[maiden ocean]

well that just reformulates it

but the general idea is that like

lets say i have a space X and a nice subspace A so we get the LES on homology

if i know H_n(A) and H_n(X, A) i can determine H_n(X) basically

or like in general exact sequences let you use information about some terms in the sequence to say things about other terms in the sequence

[dusk summit]

So i guess intuitively it provides ways to go from quotients to the space itself

[maiden ocean]

sure

or the other way around

if i know H_n(A) and H_n(X) then i can often find H_n(X/A)

[dusk summit]

thats interesting

i guess I have to get used to the idea of sequences of objects being important than, rather than just studying one object at a time

[maiden ocean]

Yeah

I mean it's kind of like how morphisms/maps end up being just as important as the objects they are between

[latent anvil]

Is there a non regular noetherian local ring R with depth 1 such that:
there is a choice of regular element x in m and y in R with m = { z in R : zy in (x) }
With the element t = y/x not satisfying any degree 2 monic polynomial

i have to do peer review and im p sure my classmate claimed something false

but somehow all the examples I can think of don't disprove it

[frank fiber]

if $H$ is a subgroup of $G_1$ and $G_2$ and $G_1/H \cong G_2/H$, then $G_1 \cong G_2$?

[cloud walrus]

Go

[viscid pewter]

no

[frank fiber]

:0

[viscid pewter]

consider H as Z2, G1 as Q8, G2 as Z2^3
then G1/Z2 iso G2/Z2 iso Z2^2, but G1 not iso G2

if it were true then life would be much nicer

[chilly ocean]

idk what Q8 is, here's another example, S3 and Z2xZ3

[viscid pewter]

q8 is the non-abelian one

but iirc classifying all of the possible Gs where G/N iso Q and you know N and Q is called like the extension problem, and it's like one of the Hard Problems in group theory

[final pasture]

G1 = Z/4Z, G2 = Z/2Z x Z/2Z and H = Z/2Z is a simpler counter-example no ?

🤔

[chilly ocean]

yes your counter-example is the best one

[viscid pewter]

lol yup, i just picked something sorta not nice that came to mind quickly so it would work nicely

[latent anvil]

Reasking my question about non regular noetherian local rings

Chmonkey pointed out that the minimal number of generators of m must be > 2. The only example I know like this is k[x, y, z, w]/(xz, xw, yz, yw) localized at (x, y, z, w) which I couldn't get to work

If you take t = x/(x-z) then t^2 = x^2/(x-z)^2 = (x^2 - xz)/(x-z)^2 = x/(x-z) = t

[next obsidian]

So

I think if you take a ring with high dimension

I feel like it’s more likely to find a counterexample?

At the very least if the dimension > 2 then the condition about m’s generators is automatically satisfied

[latent anvil]

I agree

But I'm having trouble finding examples

[latent anvil]

Let A = k[x, y, z, w]/(xz, xw, yz, yw)

tterra moment

Tfw you're a geometer but you're scared of two planes in 4d intersecting at a point

Consider the element b = (x-z)(y-w) = xy+zw

I think this is regular

I want to find an a in A such that (x, y, z, w) = { c in A : ac in (b) }

Ie such that the annihilate of a+(a) in A/(a) is m = (x, y, z, w)

This should be possible for depth reasons

a = xy

then za = wa = 0 and xa = x^2 y = x(xy + zw) = xb and ya = xy^2 = (xy+zw)y = yb

this implies m is contained in (b):a and m is maximal and (b):a is proper since a isn't in (b)

FUCK

If t = a/b then t^2 = t still

hate this

[uncut girder]

Unbased

Algebra is symbols

Hate

[latent anvil]

this is geometry @uncut girder

[uncut girder]

Oh then its good

[chilly ocean]

You can't estimate my level of thinking if you don't know even elementary terminology, so this is just emotions.

[chilly ocean]

How to show whether sqrt(5) is in Q(sqrt(2), sqrt(3))? just hint if possible

what i tried gives a horrible system of nonlinear equations that idk how to deal with

[latent anvil]

I don't think it is

[sturdy mirage]

it certainly isn't, but how to show it i'm not sure..

[latent anvil]

I think I see how

Q(sqrt(2),sqrt(3)) is galois with galois group Klein four

[solemn rain]

think of polynomials

[latent anvil]

It has exactly three intermediate degree 2 subfields

Which are Q(sqrt(2)), Q(sqrt(3)), and Q(sqrt(6))

If sqrt(5) is in there, Q(sqrt(5)) must be one of these

And this is relatively easy

[chilly ocean]

ohh

hmm

[latent anvil]

assume sqrt(5) = a + b sqrt(n) for a, b in Z and n = 2,3,6, bash

Does this outline make sense Carla?

[chilly ocean]

yea

[latent anvil]

I think it's also true that Q(sqrt(p1), sqrt(p2),...,sqrt(pn)) has degree 2^n for any distinct primes p1,...,pn

I seem to remember proving this

But I also remember it's hard

Just like, this is another more general way you could prove if

[chilly ocean]

i dont think you should say something certainly isnt true if you dont know how to prove it

[latent anvil]

eh, who cares. Maybe they're wrong and need to say mea culpa

[sturdy mirage]

why not? i could've heard a famous mathematician claim it, but not understood the proof myself

[chilly ocean]

trust noone not even yourself

[latent anvil]

hyperbole is literally the best thing ever

[sturdy mirage]

anyway, to show that its not in Q(sqrt(2)) you can check what would happen if it did. then there would exist $a,b \in \mathbb{Q}$ such that $a + b \sqrt{2} = \sqrt{5}$. or $a^2 + 2ab\sqrt{2} + 2b^2 = 5$. i dont know if its a good enough as a proof, but "certainly" it is true that either a or b has to be 0 here (to get rid of the sqrt(2) term which is irrational)

[cloud walrus]

reking

[sturdy mirage]

its not possible

and you can do something similar with Q(sqrt(3))

and 6

[latent anvil]

The thing to justify your "certainly" is that sqrt(2) and 1 are linearly independent over Q, which follows from irrationality of sqrt(2)

More directly, if ab ≠ 0 you can solve for sqrt(2) in terms of rationals

[latent anvil]

Anyone want to verify a computation

Say k is a field of char 0

let $A = k[x, y]/(xy(x+y))$

[cloud walrus]

shamrock

[latent anvil]

Let $b = x-2y, a = y^2(x+y)$

[cloud walrus]

shamrock

[latent anvil]

Is it right that $ya = -\frac{yb}{6}(3x-y)$?

[cloud walrus]

shamrock

[latent anvil]

No I have no idea how I got this

[cloud walrus]

shamrock

shamrock

[latent anvil]

,w expand (x-b)y((x-b)+y)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by shamrock#0598

[latent anvil]

Observe that x^2y + xy^2 = xy(x+y) = 0

,w b*y - 2xy - y^2 where b = (x-2y)

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by shamrock#0598

[latent anvil]

So 6y^3 = - by(x+3y)

So $ya = y^3(x+y) = -\frac{by}{6}(x+3y)(x+y)$

[cloud walrus]

shamrock

[latent anvil]

AHHH THIS EXAMPLE DOESN'T WORK

I thought for some reason the maximal ideal needs 3 generators

Which is obviously false

I'm gonna to cry

[latent anvil]

If anyone can find a counterexample for me claim your karma at mse

https://math.stackexchange.com/q/4148097/408287

[latent anvil]

Okay other counterexample: find an example with all this setup but $x \in \mathfrak{m}^2$ and $t\mathfrak{m} = R$

[cloud walrus]

shamrock

[latent anvil]

so basically find a blah blah blah ring R with a regular element x = yz where y, z in m and ((x) : (y)) = m

Hello shika

[final pasture]

Hello

[latent anvil]

I think take R = k[a, b]/(a^2 - b^3) works. Take x = a^2, then R/(a) is k[b] /(b^3), so y = b^2 works. Take t = y/x = b^2/a^2 = 1/b. Then clearly tm = 1 since it contains t * b

Smh my classmate is going to get torn to shreds on this peer review

I didn't even catch this when I read through

[latent anvil]

Nope, this isn't right

ay isn't in (x)

[chilly ocean]

cite a paper from the ussr that doesn't exist on your peer review to get around having to do any work

[chilly ocean]

да

[final pasture]

open google translate

[latent anvil]

Idk why I'm so focused on this

They didn't justify a bunch of stuff and failed to consider several cases, I already have a pretty harsh review

But I like to have counterexamples rather than just "you didn't say X"

[vestal snow]

I need help with a question from AG. Given a sheaf of sets, we know that a sheaf morphism is epic iff it is surjective on every stalk. What if it was a sheaf of rings? Would the statement be that the morphism is epic iff it is epic on every stalk?

[uncut girder]

Its epic iff it has a reputation if being cool among its peers

[next obsidian]

Yes

I believe so...

Oh fugg, I know it’s true for sheaves of abelian groups

[vestal snow]

Just proved it for sheaves of things with an underlying set

[chilly ocean]

Everytime I see your name I read Hahn Banach at first.

[vestal snow]

Wdym?

[rustic crown]

lol once more?

[hidden haven]

i mean, it was a good joke

[slender sable]

If the linear transformation T:R3→R2is defined as
T(x1,x2,x3)=(x1−x2,2x3), then the nullity of T will be what?

??

[golden pasture]

#linear-algebra

[mystic jungle]

that 7 is a typo right

what should it read instead

ah

thank you

[urban acorn]

nullT is defined as {x in R^3 | Tx = 0} so write x = (x1, x2, x3), then Tx = 0 iff (x1-x2, 2x3) = 0, which is true iff x1 = x2 and x3 = 0.

so nullT = { (x, x, 0) in R^3 | x in R }

it's really straightforward, you should practice these sort of exercises that are direct applications of the definitions

if you understand how to write proofs, and you understand the definitions, this should be easy

(Not trying to belittle you, I'm trying to point out what you should work on)

[chilly ocean]

Looks fine I think

[hidden haven]

Try doing it indirectly, by factoring x^n-1 over Q

Right, that pattern extends

x^n - 1 is the product of dth cyclotomic polynomials, where d divides n

Look at their roots, you'll see they're distributed as given in the problem

And looking at the roots is how you solve the problem

It won't do 2 directly, you'll have to use Gauss lemma

Ok let me think about this

[hidden haven]

You get this

[vestal snow]

Is this a one line proof?

[hidden haven]

But from here it's not immediate that those cyclotomics are integer

I don't see how you got it

[vestal snow]

Using that the stalk of the sheafification is naturally isomorphic to the stalk of the presheaf

[hidden haven]

Its not, I've only given the beginning of the argument

Once you have that equation, you can induct on n

Also I think I figured out this one as well

In case you'd prefer to do it that way

So you know what the automorphisms of Q(omega)/Q are?

Yes

But notice that every primitive root is a power of another

So you can give a nice description of the automorphisms

Yes

(ie they are maps which fix Q and send omega to omega^i for i = 1,...,phi(n))

So now consider any one such map, and you want to show that it fixes the polynomial's coefficients

Try proving that any such automorphism only permutes the omega_i's

ie the primitive roots map to other primitive roots, and no automorphism will map 2 to the same

Oh lol

But this specific thing just follows from the injectivity of the automorphism + the fact that roots of a polynomial map to other roots of the polynomial

But the second thing has a subtlety

Roots of a rational polynomial will map to other roots of that polynomial and you haven't proven than cyclotomic to be rational yet

But you will notice that all these omega_i's are characterized by being the nth roots of unity, but not being the kth roots of unity for any k<n

Anyway if you have proven this thing before you can use it

So if an automorphism permutes the omega_i's, and fixes Q, can you show that it fixes the given polynomial?

You don't know whether the coefficients are rational yet

That's what you have to prove

hmm I didn't get that argument

Yes

Yeah

Try to just apply the automorphism to the given version of the polynomial

Don't expand it out

(each factor won't be fixed)

Use the fact that the roots are just being permuted to say that the whole polynomial is somehow fixed

(the factors just get permuted)

t is the variable lol

You are doing this in Q(omega)[t]

Because then you don't know the coefficients

[compact needle]

It depends on how willing you are to believe that filtered colimits commute with exact sequences (ie that taking stalks on a short exact sequence of presheaves yields a short exact sequence).

[hidden haven]

They'll be complicated sums of products of omega_i's

Without expanding it's easier because you know what everything is

So do you see that when you apply an automorphism to that polynomial, all that does is permuting the factors?

Because you already know that it only permutes the roots

Yeah, if f is an automorphism, f(ab) = f(a)f(b)

You've extended whatever automorphism you have of Q(omega) to Q(omega)[t] by just having it act on each coefficient

So you get that the automorphism just permutes the factors

Then do you see that the polynomial is unchanged?

Then the coefficients are unchanged, so you're done

Yeah exactly, permuting the factors won't do anything because multiplication in a field is commutative

(or rather multiplication in the ring Q(omega)[t] is commutative)

The coefficients are fixed by every automorphism of Q(omega)/Q

And Q(omega)/Q is a Galois extension

The only things that all automorphisms fix in a galois extension are the elements of the ground field

Do you see this?

Cool

So if the coefficients are fixed by everything, they have to be in Q

Oh right second part

No

It says that irreducible and primitive in R[x] is the same as irreducible in (Frac R)[x] where R is a domain

You take the fraction field of only the coefficient ring

Well they'll be monic, so primitivity is ok

But rather than applying this statement directly, try to see when you can say that if f(x) | g(x) in Q[x], then f(x) | g(x) in Z[x]

Ah shit you might have to use the recurrence I mentioned earlier

g(x) here will be x^n - 1

Yeah

look at the roots of x^n - 1, and of the dth cyclotomic polynomial for d | n

all the roots of x^n - 1 occur in exactly one such dth cyclotomic polynomial

yep, but you get something stronger

(product of dth cyclotomic polynomials for d|n) = x^n - 1

because all the roots occur on both sides

[cloud walrus]

Moldi rocks!!!

[hidden haven]

so now try inducting on n and proving the required result

Induct on n to prove that each cyclotomic polynomial is an integer polynomial, and you know the 2 facts, that equation and the divisibility thing you get from Gauss lemma

yeah

ah

Yeah i can see that field theory (esp gal thy) proofs are way nicer but sometimes you just gotta do it (and idk if you can apply field theory here since youre working with Z)

weird, nothing particularly wrong with induction

its rather nice in its own way

you get from the first part that that equation is an equation in Q[t]

and by induction hypothesis the product of everything on the left, except the nth cyclotomic, has integer coefficients and is monic

(therefore primitive)

so you have f(t)(nth cyclotomic) = x^n - 1 where f(t) is primitive

use the fact that if f(t) | g(t) in Q[t] and both are integer polynomials, and f(t) is primitive, then divisibility holds in Z[t] as well

then use uniqueness of factorization in Q[t]

induction hypothesis

the dth cyclotomic polynomials for d<n are all integer polynomials

and you want to prove that the nth one is too

f(t) in my notation was the product of the cyclotomic polynomials with d | n, d<n

[cloud walrus]

Moldi rocks!!!

[hidden haven]

I just wrote this as $f(x)\Phi_n(x) = x^n - 1$ where $f(x) = \prod_{d|n, d<n}\Phi_d(x)$

[cloud walrus]

Moldi rocks!!!

[hidden haven]

f(x) itself is not cyclotomic

it is a product of cyclotomic polynomials

d can be equal to n

but yes

and d=1 as well

[hidden haven]

hence this equation

@vestal snow channel free now

[vestal snow]

Can someone verify my proof that cokernel of stalks are isomorphic to stalks of cokernel?

CamScanner 05-23-2021 19.05

Is it okay so far?

The last claim seems to be a diagram chase so I haven't done that yet

[vestal snow]

Is this asking me to prove that $Ker_{Sh}((Coker_{Psh}(\phi))^{Sh})$ is isomorphic to $(Ker_{Psh}(Coker_{Psh}(\phi)))^{Sh}$?

[cloud walrus]

Hahn Banach

[compact needle]

I think this is good, other than I would make a point to say the reason that \varphi_V exists is because the image of F(V) in coker(\phi_p) must be zero by an appropriate diagram chase, hence you can use the universal property of coker\phi(V).

[compact needle]

Yeah, I read the question the same way you do.

[vestal snow]

Thanks

How do I show that the tensor product of stalks is the stalk of the tensor product?

[sturdy marsh]

left adjoints commute with colimits

[vestal snow]

Won't I have to prove first that the sheaf tensor product is a left adjoint?

[sturdy marsh]

yes

[vestal snow]

Okay assume that we know that the sheaf tensor is left adjoint

How does it follow from that?

[sturdy marsh]

actually you dont need the sheaf tensor

the stalk of the tensor product sheaf is the stalk of the tensor product presheaf (i.e. just tensor everything levelwise)

and then use the fact that tensor product of modules commutes with colimits

Colim (M_i \otimes N_i) = Colim(M_i) \otimes (N_i)

[vestal snow]

There's something about this that I think does not work

At each level, we are tensoring with different rings

[sturdy marsh]

what category are you working in

[vestal snow]

O_X modules

Where O_X is a ringed space

[sturdy marsh]

okay say you have O_X modules F and G